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Index Laws and Equations with Indices

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AQA Level 2 Further Maths practice questions, each with a hint and a full worked solution.

Have a pen and paper handy, to work out your answers.

Question 12 marks

Write \dfrac{(2x)^3\sqrt{x}}{16x^5} in the form ax^n, where a and n are constants.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Cube the 2 as well as the x, and write \sqrt{x} as x^{\frac12}.

Worked solution
  1. (2x)^3=8x^3 (cube the 2 too)
  2. \sqrt{x}=x^{\frac12}, so the top is 8x^3\times x^{\frac12}=8x^{\frac72}
  3. Divide the numbers: 8\div16=\frac12
  4. Subtract the powers: \frac72-5=-\frac32
  5. Answer: \frac12x^{-\frac32}

Question 22 marks

Solve 25^{x+1}=\dfrac{1}{5^{x}}

Give your answer as a fraction.

Hint

Write both sides as a power of 5, remembering that \frac{1}{5^x}=5^{-x}.

Worked solution
  1. 25=5^2, so 25^{x+1}=5^{2(x+1)}=5^{2x+2}
  2. \dfrac{1}{5^x}=5^{-x}
  3. Equate the powers: 2x+2=-x
  4. 3x=-2
  5. Answer: x=-\frac23

Question 3Challenge5 marks

Here are two equations.

\frac{4^{x}}{2^{y}}=32 \qquad\qquad 27^{x}\times3^{y}=9\sqrt{3}

(a)

Write the first equation as a linear equation in x and y.

1 mark

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

(b)

Write the second equation as a linear equation in x and y.

2 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

(c)

Hence work out the values of x and y.

Give your answer in the form (x,\ y).

2 marks

Write your answer as (x, y)

Hint

Write every number as a power of 2 (first equation) or a power of 3 (second), including \sqrt3=3^{\frac12}, then equate the powers.

Worked solution

Part (a)

  1. 4^x=(2^2)^x=2^{2x} and 32=2^5
  2. \dfrac{2^{2x}}{2^y}=2^{2x-y}
  3. Equate the powers of 2: 2x-y=5

Part (b)

  1. 27^x=(3^3)^x=3^{3x}, so the left side is 3^{3x}\times3^y=3^{3x+y}
  2. 9\sqrt3=3^2\times3^{\frac12}=3^{\frac52}
  3. Equate the powers of 3: 3x+y=\frac52
  4. (or 6x+2y=5)

Part (c)

  1. Add the two equations: (2x-y)+(3x+y)=5+\frac52
  2. 5x=\frac{15}{2}, so x=\frac32
  3. Substitute into 2x-y=5: 3-y=5, so y=-2
  4. Check: \dfrac{4^{1.5}}{2^{-2}}=8\times4=32 ✓
  5. Answer: x=\frac32,\ y=-2

More on this topic: Index Laws and Equations with Indices worksheet with full solutions

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